Problem: Simplify and expand the following expression: $ \dfrac{2r - 4}{r + 9}+\dfrac{r}{3r - 3} $
Explanation: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(r + 9)(3r - 3)$ Multiply the first term by $\dfrac{3r - 3}{3r - 3}$ $ \begin{align*} \dfrac{2r - 4}{r + 9} \times \dfrac{3r - 3}{3r - 3} & = \dfrac{(2r - 4)(3r - 3)}{(r + 9)(3r - 3)} \\ & = \dfrac{6r^2 - 18r + 12}{(r + 9)(3r - 3)}\end{align*} $ Multiply the second term by $\dfrac{r + 9}{r + 9}$ $ \begin{align*} \dfrac{r}{3r - 3} \times \dfrac{r + 9}{r + 9} & = \dfrac{(r)(r + 9)}{(3r - 3)(r + 9)} \\ & = \dfrac{r^2 + 9r}{(3r - 3)(r + 9)}\end{align*} $ Now we have: $ = \dfrac{6r^2 - 18r + 12}{(r + 9)(3r - 3)} + \dfrac{r^2 + 9r}{(3r - 3)(r + 9)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{6r^2 - 18r + 12 + r^2 + 9r}{(r + 9)(3r - 3)} $ $ = \dfrac{7r^2 - 9r + 12}{(r + 9)(3r - 3)}$ Expand the denominator: $ = \dfrac{7r^2 - 9r + 12}{3r^2 + 24r - 27}$